1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

5.5 • INVERSE TRICONOMETRJC AND HYPERBOLIC FUNCTIONS 191

and the corresponding value of the derivative is given by

1.

f I ( vz) = [ 1 - ( ~) 2)!

= "'." = -i.

i

The inverse hyperbolic functions a.re

arcsinh z = log [z + (z^2 +1)!],

arccosh z = log [z + (z^2 - 1) ~) , and

arctanh z = ~ log G ~ : ).

Their derivatives are

d .nh 1

dz arcs1 z = ----, ,

d

dz arccosh z =

d

(z^2 + 1p

1

(z^2 - 1)~'

1

• dz arctanh z = 1 - z (^2).
  and
  To establish Identity (5-48), we start with w = arctanh z and obtain
  ew - e-w e2w - 1
  z = tanh w = ----
  ew + e-w e2w + 1 '
  (5-48)
  which we solve for e^2 w, getting e^2 w =
  1
  1

• z. Taking the logarithms of both sides

• z
  gives

w=arctanhz=^1 (l+z)
2 log l-z,

which is what we wanted to show.

• EXAMPLE 5. 13 Calculation reveals that
  . 1 1+1+2i 1.
  arctanh(1+ 2i) =

2

tog

1

_1_

2

i =2 log(- 1 +i)

= ~ln2+i(~+n)1T,

where n is an integer.